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Stimulated emission. Energy of the photon inducing the emission.

  1. Mar 21, 2008 #1
    In order to produce stimulated emission we need a photon which has the same energy as the difference is the lower and upper energy levels in the excited atom. But how exactly the energy of the photon does have to correspond to the energy difference between the levels. Can stimulated emission still occur if the photon has little less or little more?

    If so, where does the rest of the energy come from or where does the rest of the energy go?
    Can we get the extra energy from or put the rest in some form of collisional energy?

    If we have a photon that does not exactly match for the energy of the energy levels will the stimulated emission then be only less probable according to the stimulated emission cross section.

    Einstein's coefficient for stimulated emission should in this case only be for a photon of the exact energy?
  2. jcsd
  3. Mar 21, 2008 #2
    Energy levels in atoms aren't exactly discrete, they do have some finite width in energy due to the time-energy uncertainty principle. The natural width is pretty small, but photons with slightly different energy can be absorbed by an atom.

    Yes, that is true, the absorbtion cross section is a lorentzian function. It is peaked at the wavelength corresponding to the energy difference, then falls of on either side.
  4. Mar 23, 2008 #3
    So how large will the linewidth be due this uncertainty principle? I guess that it will be quite insignificant for example when compared to linewidth caused by Doppler-effect.

    And what about this Stark effect line broadening? What is the process behind this broadening?
  5. Mar 23, 2008 #4
    Yes, the natural width due to the unceratinty relations is really small, much less than the effect of the thermal motions. For thermal broadening the width is,

    [tex] \frac{2 \lambda}{c}\sqrt\frac{2kTln(2)}{m} [/tex]

    So it depends on the mass and wavelength if the line, but say ~0.1 angstroms is in the ballpark

    and for natural broadening the width is

    [tex] \frac{\lambda^2}{\pi c t_0} [/tex]

    where t_0 is the lifetime, about 10^(-8) seconds or so. So this gives a natural line width of ~10^-4 angstroms.

    The stark effect is an effect of nearby atoms, the elcetric field of at ion or atom slghtly perturbs the energy levels of the atoms. In most cases it is pretty small, unless the density of particles is high. I can't remember the numebers off hand, but I'm pretty sure that it is more than natural and less than thermal in most cases.

    doppler > collisional > natural (for most cases . . .)
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